Problem equivalent to "largest square in a cube"

Question

The "largest square in a cube" problem, which asks for the largest square inside a cube, has a solution as can be seen on this page, which also says that the general problem in higher dimensions is unsolved.

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          ^{MathWorld image.}

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The problem is equivalent to the following optimization problem: find two orthogonal unit vectors in $\mathbb{R}^3$ such that the maximum of the absolute values of all their coordinates is minimized. For the "largest square in a cube" problem to have the answer it does, this minimum should be $2/3$, i.e., it occurs when the coordinates are $(2/3,2/3,1/3),(1/3,-2/3,2/3)$ (or some coordinate permutation-cum-axis reflection of those). How would we show that this is indeed where the optimum occurs? This seems like it should be some really simple algebra, but it is eluding me.

More abstractly, this is asking for a pair of orthogonal unit vectors such that the maximum of their $\ell^\infty$-norms is minimized. What happens if we are trying to minimize the larger of the $\ell^p$-norms, $p > 2$? Does the minimizing value tend to $2/3$ as $p \to \infty$?

Answer 1

When studying this problem in the general case, i.e. $m$-dimensional cube inside $n$-dimensional cube for $m<n$, it might make sense to look at small examples to gain some intuition. From the small examples (largest square in (hyper)cubes) one could get the impression that the coordinates of the optimal solution are always rationals and hence the optimal edge length are always square root of rationals. In this post I argue that this is not the case in general.

Here is how your formulation as an optimization problem generalizes: Let $S$ be a set of half of the $2^m$ vertices of the m-cube of diagonal lenght 2 centered at the origin such that $S \cup \{-s : s\in S\}$ is the entire set of vertices. Now find $2^{m-1}$ unit-vectors in $\mathbb{R}^n$ such that each pair of them has the same vector product as a corresponding pair in $S$, and such that the absolute value of their coordinates in minimized.

For example for m=3 one can take as $S$ every other vertex of the cube, such that they form a regular tetrahedron (see picture by Kepler), and then for each pair there vector product would have to be $-\frac{1}{3}$

One potential problem with this formulation as optimization problem in practice is that the number of variables is $n2^{m-1}$ and there are $\frac{2^{m-1}(2^{m-1}+1)}{2}$ quadratic equations.

Together with this optimization formulation and methods described in http://arxiv.org/abs/1407.0683 I calculate some cases $f(m,n)$ for small $m$ and $n$. There is some numerical optimization involved, so it is not a formal proof that the following result is optimal, but I obtain exact symbolic values for the coordinates and it is certainly a lower bound.

On mathworld the cases $f(1,n)$ and $f(2,n)$ are explained, so the smallest cases that seem to be unknown are $f(3,n)$. The cases $f(3,4)$ and $f(3,5) $ are also mentioned in Sloane's oeis: A243309, A243313 .

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• $f(3,4)$:

The 8 vertices of a largest 3-cube inside $[0,1]^4$ have the following coordinates:

$$(1, 0, 1-a, b),(1, 0, b, 1-a), (0,c,1-d, 0), (0,c, 0, 1-d),(1, 1-c, 1,d), (1, 1-c,d, 1), (0,1, 1-b, a),(0, 1, a,1-b)$$

where $a,b,c,d$ are algebraic numbers of degree $4$, with the these minimal polynomials and decimal approximations:

$$\begin{align*}a\quad&16x^4 + 8x^3 - 23x^2 + 14x - 2& 0.204901553506651293143\\ b\quad&16x^4 - 24x^3 + 25x^2 - 14x + 1& 0.082734498297453867827\\ c\quad&8x^4 - 32x^3 + 45x^2 - 30x + 1& 0.035139649420907685891\\ d\quad&4x^4 - 4x^3 - x^2 + 4x - 1& 0.287636051804105160970 \end{align*}$$

This gives a edge length s, which hast the following minimal polynomial:

$$4x^8 - 28x^6 - 7x^4 + 16x^2 + 16$$

and $1.007434756884279376098253595231$ as decimal approximation.

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• $f(3,5)$:

The 8 vertices of a largest 3-cube inside $[0,1]^5$ have the following coordinates: $$(1, e, 0, e, 0), (1, e, 0, 1, 1-e), (0, 0, e, 0, e), (0, 0, e, 1-e, 1), (1, 1, 1-e, e, 0), (1, 1, 1-e, 1, 1-e), (0, 1-e, 1, 0, e), (0, 1-e, 1, 1-e, 1)$$

where $e=\sqrt{\frac{3}{2}}-1\approx 0.224744871391589049098$

This gives as edge length $\sqrt{11-8\sqrt{\frac{3}{2}}}\approx 1.09637631717731280407593110$.

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• $f(3,6)$:

The 8 vertices of a largest 3-cube inside $[0,1]^6$ all lie on the vertices of (a diagonal section of) the 6-cube:

$$(0, 1, 1, 0, 1, 0), (0, 0, 0, 1, 1, 1), (1, 0, 1, 0, 0, 1), (1, 1, 0, 1, 0, 0), (1, 0, 0, 1, 0, 1), (1, 1, 1, 0, 0, 0), (0, 1, 0, 1, 1, 0), (0, 0, 1, 0, 1, 1)$$

and this gives an edge length of $\sqrt{2}\approx 1.414213562373095048$.

Answer 2

You can reduce things a bit as follows. Suppose your two vectors are (a,b,c) and (x,y,z). Without loss of generality all of a, b and c are non-negative. Let's suppose that a is bigger than both b and c. Pick a third vector (u,v,w) orthogonal to both (a,b,c) and (x,y,z). If u is non-zero, then we can add a tiny multiple of (u,v,w) to (a,b,c) in such a way that the $\ell_2$ norm increases by a quadratic amount but the $\ell_\infty$ norm decreases by a linear amount. So after rescaling we have got a better example.

I won't go through all possible cases here, but we now know that either u is zero or (WLOG) a=b. The same proof also tells us that the maximum absolute value of x, y and z is attained twice. Probably it's fairly easy to show, after a little bit of case analysis, that this maximum is attained once for a positive value and once for a negative value, and in places 2 and 3. If that works, then we've got to (a,a,b) and (x,-y,y), with a bigger than b and x bigger than y in absolute value. This starts to look like a much more manageable optimization problem.

I don't guarantee that all this works, but I think it has a good chance.